http://gcc.gnu.org/bugzilla/show_bug.cgi?id=47004
Geert Bosch <bosch at gcc dot gnu.org> changed:
What |Removed |Added
----------------------------------------------------------------------------
CC| |bosch at gcc dot gnu.org
--- Comment #1 from Geert Bosch <bosch at gcc dot gnu.org> 2010-12-18 19:22:13
UTC ---
The compiler gets really close in being able to optimize this testcase.
In fact, adding an extra always-true expression, as I did in lenzero4,
causes the compiler to optimize correctly.
int lenzero3 (int f, int l) {
return (l < f ? 0 == 0 : l - f + 1 == 0);
}
int lenzero4 (int f, int l) {
return (l < f ? 0 == 0 : (l - f >= 0) && (l - f + 1 == 0));
}
The first generates:
_lenzero3:
0000000000000040 cmpl %edi,%esi
0000000000000042 movl $0x00000001,%eax
0000000000000047 jl 0x00000053
0000000000000049 subl %edi,%esi
000000000000004b xorl %eax,%eax
000000000000004d cmpl $0xff,%esi
0000000000000050 sete %al
0000000000000053 repz/ret
The second:
_lenzero4:
0000000000000060 xorl %eax,%eax
0000000000000062 cmpl %edi,%esi
0000000000000064 setl %al
0000000000000067 ret
The extra condition (l - f >= 0), provides the compiler with the right
hint to do the optimization. It can prove this added condition is always
true, and it then can derive that the condition (l - f + 1 == 0) is
always false.
So, the compiler *can* prove l < f implies l - f >= 0, it just doesn't
know it should prove it in order to optimize the expression in lenzero3.
-Geert
